Apply qualitatively and quantitatively Newton's Law of Universal Gravitation, F = G m_1 m_2 / r^2, to determine the magnitude of force, gravitational field strength g = G M / r^2, and acceleration due to gravity at different points in a radial gravitational field

What this dot point is asking

NESA wants you to apply Newton's Law of Universal Gravitation to calculate the force between two masses, determine the gravitational field strength at a point, and explain how the inverse-square dependence on distance shapes planetary and satellite motion. You need both the conceptual explanation (action at a distance, field model) and the numerical fluency to compute $F$ and $g$ at arbitrary distances.

The answer

Newton's Law of Universal Gravitation

Every pair of point masses attracts each other with a force directed along the line joining them:

where:

• IMATH_4 N m$^2$/kg$^2$ is the universal gravitational constant.
• IMATH_7 and $m_2$ are the two masses in kilograms.
• IMATH_9 is the distance between their centres (not surfaces).

The force is mutual: each mass exerts the same magnitude of force on the other (Newton's third law).

The inverse-square law

Force is inversely proportional to the square of the distance. Doubling $r$ reduces the force to one quarter. Halving $r$ quadruples the force. This rapid fall-off explains why Earth's gravity dominates near the surface but becomes negligible far from the planet.

Gravitational field strength

The gravitational field strength $g$ at a point is the gravitational force per unit mass on a test mass placed there:

where $M$ is the mass of the source body and $r$ is the distance from its centre. Units: N/kg or m/s$^2$ (numerically equal).

At Earth's surface ($r = R_E = 6.37 \times 10^6$ m): $g \approx 9.8 \text{ m/s}^2$.

Acceleration due to gravity

For an object of mass $m$ in free fall in a gravitational field $g$, the acceleration is $a = g$ (regardless of $m$, because $F = mg$ and $a = F/m$). All objects fall with the same acceleration in a given gravitational field, in the absence of air resistance.

Field model versus action at a distance

Two equivalent descriptions:

• Action at a distance: the two masses pull on each other directly across empty space.
• Field model: each mass creates a gravitational field around itself, and any other mass in that field experiences a force. The field model is preferred for HSC because it generalises cleanly to electric and magnetic fields.

Worked example with numbers

Calculate the gravitational force between the Earth ($M_E = 5.97 \times 10^{24}$ kg) and the Moon ($M_m = 7.35 \times 10^{22}$ kg) separated by $r = 3.84 \times 10^8$ m.

$F = G \frac{M_E M_m}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^8)^2}$

Numerator: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22} = 2.93 \times 10^{37}$.

Denominator: $(3.84 \times 10^8)^2 = 1.47 \times 10^{17}$.

$F = \frac{2.93 \times 10^{37}}{1.47 \times 10^{17}} = 1.99 \times 10^{20}$ N.

This is the force that holds the Moon in orbit around the Earth.

Try it: Universal gravitation calculator - plug in any two masses and separation to get $F$ and $g$, with Earth-Moon and Sun-Earth presets.

Common traps

Using altitude instead of distance from the centre. $r$ in the formula is always measured from the centre of the source body. For a satellite at altitude $h$ above Earth: $r = R_E + h$.

Forgetting to square $r$. The denominator is $r^2$, not $r$. Halving the distance multiplies the force by four, not two.

Confusing $g$ and $G$. $G$ is a universal constant ($6.67 \times 10^{-11}$). $g$ depends on the source mass and your distance from it.

Assuming $g = 9.8 \text{ m/s}^2$ everywhere. This is only true at Earth's surface. At higher altitudes or on other planets, recalculate using $g = G M / r^2$.

Treating gravity as having a cut-off. Gravity extends to infinity, just very weakly. The "edge" of Earth's gravity is fictional.

In one sentence

Newton's law of universal gravitation, $F = G m_1 m_2 / r^2$, gives the attractive force between any two masses, and the resulting field strength $g = G M / r^2$ falls off as the inverse square of the distance from the centre of the source.